### Video Transcript

Discuss the continuity of the
function π at π₯ is equal to zero, given that π of π₯ is equal to negative five
sin of eight π₯ over π₯ plus eight multiplied by the cos of π₯ if π₯ is not equal to
zero and π of π₯ is equal to negative 32 if π₯ is equal to zero.

The question wants us to determine
the continuity or discontinuity of our piecewise function π at the point π₯ is
equal to zero. We recall we call a function π
continuous at the point π₯ is equal to π if the following three conditions are
true. First, we need that π evaluated at
π is defined. This is the same as saying that π
is in the domain of our function π. Second, the limit as π₯ approaches
π of π of π₯ must also exist. And we recall this is equivalent to
saying the limit as π₯ approaches π from the right of π of π₯ and the limit as π₯
approaches π from the left of π of π₯ both exist and are equal. Finally, we must have the limit as
π₯ approaches π of our function π of π₯ is equal to π evaluated at π.

So to check our piecewise function
π is continuous at π₯ is equal to zero, we need to check all three of these
conditions are true. So weβll set π equal to zero and
check each of these conditions in turn. Weβll start by checking that π
evaluated at zero is defined. To evaluate π at zero, we see that
our function is defined piecewise. And we see when π₯ is equal to
zero, our function is defined to output negative 32. So π evaluated at zero is equal to
negative 32. And that means that π evaluated at
zero is defined. So our first condition is true.

So weβll now move on to the second
condition, checking that the limit as π₯ approaches zero of our function π of π₯
exists. To do this, we need to check the
limit as π₯ approaches zero from the right of π of π₯ and the limit as π₯
approaches zero from the left of π of π₯ both exist and are equal. Weβll start with the limit as π₯
approaches zero from the right. Since π₯ is approaching zero from
the right, we have that π₯ is greater than zero. And we see from our piecewise
definition of the function π of π₯, when π₯ is not equal to zero, our function π
of π₯ is equal to negative five sin of eight π₯ over π₯ plus eight cos of π₯.

So since π₯ is approaching zero
from the right, π₯ is never equal to zero. And from our piecewise definition
of π of π₯, we have that π of π₯ is equal to this expression whenever π₯ is not
equal to zero. This means their limits will be
equal. We might then try direct
substitution. However, since π₯ is approaching
zero, we see that five sin of eight π₯ is approaching zero and π₯ is approaching
zero. This gives us zero over zero, which
is an indeterminate form. This means weβre either going to
have to manipulate our limit into one we can evaluate or weβre going to need to use
our standard trigonometric limit laws.

Since our expression contains the
sine function divided by π₯, we recall that the limit as π₯ approaches zero of the
sin of π₯ divided by π₯ is equal to one. This is a standard trigonometric
limit result which we should commit to memory. However, in our limit, we have the
sin of eight π₯ in our numerator. So weβll replace all instances of
π₯ with eight π₯ in our standard limit result. We see that if eight π₯ is
approaching zero, then π₯ must be approaching zero. And we see that one-eighth is a
constant, so we can take this constant outside of our limit. Finally, we can multiply both sides
of this equation by eight, giving us the limit as π₯ approaches zero of the sin of
eight π₯ divided by π₯ is equal to eight.

Now, by using the fact that a limit
of a sum is equal to the sum of a limit, we can split our limit into the sum of
these two limits. And since negative five and eight
are constants with respect to π₯, we can take these coefficients outside of our
limit. Weβre now ready to evaluate these
limits. Weβve already shown that the limit
as π₯ approaches zero of the sin of eight π₯ divided by π₯ is equal to eight. And since the cos of π₯ is a
standard trigonometric function, we can evaluate this by using direct
substitution. Substituting π₯ is equal to zero
gives us the cos of zero, which we can evaluate to give us one. Therefore, our limits are equal to
negative five times eight, which is negative 40, plus eight times one, which is
equal to eight. This simplifies to give us negative
32.

So weβve shown that our limit as π₯
approaches zero from the right of π of π₯ is equal to negative 32. We now need to show that the limit
as π₯ approaches zero from the left of π of π₯ is also equal to negative 32. Letβs see what wouldβve happened if
instead of taking the limit as π₯ approaches zero from the right using our working,
weβd instead use the limit as π₯ approaches zero from the left. Well, since π₯ is approaching zero
from the left, we now have that π₯ is less than zero. However, we only use the fact that
π₯ was not equal to zero in our working. So in our left-hand limit, our
function π of π₯ is still equal to this expression.

In fact, none of the steps in our
working specifically used the fact that we were taking a right-hand limit. Every step in our working will be
exactly the same when weβre taking the left-hand limit. This means we can conclude that the
limit as π₯ approaches zero from the left of π of π₯ is also equal to negative
32. So since both the left-hand and
right-hand limit exist and are equal, weβve shown that the limit as π₯ approaches
zero of π of π₯ exists.

The final thing we need for our
function π of π₯ to be continuous at zero is the limit as π₯ approaches zero of π
of π₯ is equal to π evaluated at zero. We already found both of these
values when checking the first and second conditions. When we checked the first
condition, we showed that π evaluated at zero is equal to negative 32. And when we checked our second
condition, we showed that the limit as π₯ approaches zero from the left and the
limit as π₯ approaches zero from the right of π of π₯ is equal to negative 32. This is the same as saying the
limit as π₯ approaches zero of π of π₯ is equal to negative 32. So we have the limit as π₯
approaches zero of π of π₯ is equal to negative 32, which is also equal to π
evaluated at zero. So our third condition for
continuity is also true.

Therefore, since our function π of
π₯ satisfied all three conditions for continuity at π₯ is equal to zero, we can
conclude that our function π is continuous at π₯ is equal to zero.